If we have an $n$-sided regular polygon with side lengths of $a$, what is the area consisting of all the points whose distances to the center of the polygon are shorter than their distances to the perimeter?
Below is a diagram describing my question when $n$ = 4. (The shaded part is the area we are trying to find)
$n$ = 4" />I am familiar with how I can solve for the square case (it was answered here), however I am unsure how we can extend this to the general case for a regular polygon with $n$ sides?
Any help with this question would be greatly appreciated.
COMMENT.-The attached figure corresponds to a regular heptagon inscribed in a circle of radius $5$ (that is, your $a=10\sin\left(\dfrac{\pi}{7}\right)$). We first determine the locus of points whose distances to the origin and to the side $a$, are equal, with which the delimiting border of the searched area, say $A_0$, is found and the total area is in this case $7A_0$. Clearly the same is valid for the central angle $\dfrac{2\pi}{n}$ in the general case of an $n$-gon.
The triangle of reference, put in the position shown in the figure, let $P=(x,y)$ a point inside the triangle, equidistant of the origin and the side $a$. The corresponding locus is easily found and the curve solution is an arc of a kind of parabola (black curve in the figure). The area $A_0$ is the shaded region.
Having the line $CB$ by $Ax+By+C=0$ and $P=(x,y)$ the condition of said equidistance is given by $$\sqrt{x^2+y^2}=\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}\tag 1$$ which is just the equation of the locus.
The calculation of the area $A_0$ could perhaps be facilitated by going in $(1)$ to polar coordinates, and integrating from $0$ to $\dfrac{2\pi}{n}$.